Question

suppose that the number of bacteria in a certain population increases according to a continuous exponential growth model. a sample of 2700 bacteria selected from this population reached the size of 2946 bacteria in two hours. find the hourly growth rate parameter. note! this is a continuous exponential growth model. write your answer as a percentage. do not round any intermediate computations, and round your percentage to the nearest hundredth

Explanation:

Step1: Recall the continuous - exponential growth model

The formula for continuous - exponential growth is \(P(t)=P_0e^{rt}\), where \(P(t)\) is the population at time \(t\), \(P_0\) is the initial population, \(r\) is the growth rate parameter, and \(t\) is the time.
We are given that \(P_0 = 2700\), \(P(t)=2946\) when \(t = 2\).
Substitute these values into the formula: \(2946=2700e^{2r}\).

Step2: Solve for \(r\)

First, divide both sides of the equation \(2946 = 2700e^{2r}\) by \(2700\):
\(\frac{2946}{2700}=e^{2r}\).
Simplify \(\frac{2946}{2700}=1.09111\cdots\), so \(1.09111\cdots=e^{2r}\).
Take the natural logarithm of both sides: \(\ln(1.09111\cdots)=\ln(e^{2r})\).
Since \(\ln(e^{x})=x\), we have \(\ln(1.09111\cdots)=2r\).
We know that \(\ln(1.09111)\approx0.087\).
Then \(r=\frac{\ln(1.09111)}{2}\).

Step3: Calculate the percentage

To convert \(r\) to a percentage, we use the formula \(r(\text{percentage})=r\times100\).
Since \(r=\frac{\ln(1.09111)}{2}\approx\frac{0.087}{2}=0.0435\), then \(r(\text{percentage})=0.0435\times 100 = 4.35\%\)

Answer:

\(4.35\)